It is best if we get this into the form of #(1+ax)^n# To do so, we must multiply everything in the brackets by #x^2#, which is the same as dividing by #1/x^2#.
#x^2(1/x^2+3x)=1+3x^3#
Step 1:
#(1/x^2+3x)^6=(1/x^2(1+3x^3))^6#
#=(1/x^2)^6(1+3x^3)^6#
#=1/x^12(1+3x^3)^6#
Step 2:
We know that a version of binomial expansion is:
#(1+x)^n=1+nx+(n(n-1)(x)^2)/(2!)+(n(n-1)(n-2)(ax)^3)/(31)+cdots+n(x)^(n-1)+(x)^n#
We will set aside our #1/x^12# for later.
We can substitute #3x^3# for #x#, and #6# for #n#:
#(1+3x^3)^6=1+6(3x^3)+(6(5)(3x^3)^2)/(2!)+(6(5)(4)(3x^3)^3)/(3!)+(6(5)(4)(3)(3x^3)^4)/(4!)+6(3x^3)^5+(3x^3)^6#
#=1+18x^3+135x^6+540x^9+1215x^12+1458x^15+729x^18#
Step 3:
We now need to multiply each term by #1/x^12# since #1/x^12(1+3x^3)^6=(1/x^2(1+3x^3))^6=(1/x^2+3x)^6#
#1/x^12(1+18x^3+135x^6+540x^9+1215x^12+1458x^15+729x^18)#
#=1/x^12+18/x^9+135/x^6+540/x^3+1215+1458x^3+729x^6#
By factorising:
#=1/x^12+18/x^9+135/x^6+540/x^3+1215+729x^3(2+x^3)#
